overviews of statics
TRANSCRIPT
L E C T U R E R : P U I P U I L E E
B A S I C T E X T:1 . R . C . H I B B E L E R , E N G I N E E R I N G M E C H A N I C S : S TAT I C S,
1 2 T H E D. , N E W J E R S E Y, P R E N T I C E H A L L , 2 0 0 7 .
R E F E R E N C E S :1 . M E R I A M , J. L . , E N G I N E E R I N G M E C H A N I C S, VO L 1
S TAT I C S, 6 T H E D ( S I E D I T I O N ) . , J O H N W I L E Y & S O N S, 2 0 0 8 . 2 . A N D R E W P Y T E L A N D J A A N K I U S A L A A S. E N G I N E E R I N G
M E C H A N I C S : S TAT I C S 2 N D E D. T H O M S O N L E A R N I N G, 2 0 0 1 .
EGR1174- Engineering Mechanics:
Statics
Overview
1. General Principles & definitions
2. Fundamental concepts
3. Units of measurement, international systems of units
4. Scalars and Vectors
1. General Principles & Definition
Statics and Dynamics are introductory engineering mechanics courses, and they are among the first engineering courses encountered by most students.
Therefore it is appropriate that we begin with a brief exposition on the meaning of the term “engineering mechanics”
1. General Principles & Definition
‘Physics’ – the science that relates the properties of matter and energy, excluding biological and chemical effects. Physics includes the study of mechanics, thermodynamics, electricity and magnetism, and nuclear physics.
Mechanics’ – the branch of physics that considers the action of forces or fluids that are at rest or in motion. Correspondingly, the primary topics of mechanics are statics and dynamics.
Physics
Thermodynamics
Mechanics Electricity
Deformable-Body Rigid-Body
Fluid
Statics Dynamics
Magnetism
1. General Principles & Definition
‘Engineering’- the application of the mathematical and physical sciences (physics, chemistry and biology) to the design and manufacture of items that benefit humanity.
‘Engineering Mechanics’ – the branch of engineering that applies the principles of mechanics to mechanical design (i.e. any design that must take into account the effect of forces)
Introduction to statics
Mechanics is the oldest of the physical sciences.
The earliest recorded writings in mechanics are those of Archimedes (287-212 B.C.) on the principle of the lever and the principle of buoyancy.
Substantial progress came later with the formulation of the laws of vectors combination of forces by Stevinus (1548-1620), who also formulated most of the principles of statics.
Fundamental Concept
Four basic quantities are used throughout mechanics
1. Length- used to locate the position of a point in space and thereby describe the size of a physical system. It can be used to define distances and geometry
2. Time- the measure of the succession of events. Although the principles of statics are time independent, this quantity plays an important role in Dynamics
3. Mass- a measure of the inertia of body, which is its resistance to the change of velocity. Mass can also be thought of as the quantity of matter in a body. The mass of a body affects the gravitational attraction force between it and other bodies. This force appears in many applications in statics
4. Force- the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterised by its magnitude, by the direction of its action, and by its point of application. Force is a vector quantities.
Fundamental Concepts
3 Idealisation are used in mechanics in order to simplify application of the theory.
1. Particle- a particle has a mass, but a size that can be neglected. 1. i.e. The size of the earth is insignificant compared to the size of its orbit, thus the size of the earth
can be modeled as a particle when studying its orbital motion. 2. When a body is idealised as a particle, the principles of mechanics reduce to a rather simplified form
since the geometry of the body will not be involved in the analysis of the problem
2. Rigid Body- a rigid body can be considered as a combination of a large number of particles in which all the particles remain at a fixed distance from one another, both before and after applying a load.
1. This is important because the material properties of any body that is assumed to be rigid will not have to be considered when studying the effects of forces acting on the body. In most cases, the actual deformations occurring in structures, machines, mechanisms are relatively small, thus the rigid-body assumption is suitable for analysis
3. Concentrated Force- it represents the effect of a loading which is assumed to act at a point on a body.
1. We can represent a load by a concentrated force, provided the area over which the load is applied is very small compared to the overall size of the body. I.e. the contact force between a wheel and the ground.
2.Fundamental Concepts
Engineering mechanics is formulated on the basis of Newton’s three laws of motion.
First law. Particle originally at rest, or moving in a straight line with constant velocity,
tends to remain in this state provided the particle is not subjected to an unbalanced force.
Second law. A particle acted upon by an unbalanced force F experiences an acceleration a
that has the same direction as the force and a magnitude that is directly proportional to the force.
Third law. The mutual forces of action and reaction between two particles are equal,
opposite, and collinear.
2.Fundamental concept
Shortly after formulating his 3 law, Newton postulated a law governing the gravitational attraction between any two particles.
F = Gm1m2/r² G= universal constant gravitation, according to experimental evidence,
G=66.73X10^-12 m³/(kgs²) m1 m2 – mass of each of the two particles r= distance between the two particles
W= Gm1Me/r² m1 = weight of a particle assume earth to have constant density and mass of m2=Me r = distance between the earth’s center and the particle
W=mg By comparison with F=ma, we can see that g is the acceleration due to gravity.
Since it depends on r, then the weight of a body is not an absolute quantity. Instead, its magnitude is determined from where the measurement was made. For most engineering calculations, however, g is determined at sea level and at a latitude of 45º, which is considered the ‘standard location’
3. Units of measurement
The four basic quantities are related by Newton’s second law of motion, F=ma
SI units. International Systems of Units defines length in Meters (m), time in Seconds (s), mass in Kilograms (kg), force in Newton (N).
F= ma, = kg (m/s²) = N
SI symbols and Prefixes: 1000 k, kilo 1000 000 M, mega 1000 000 000 G, giga 0.001 m, milli 0.000 001 µ, micro 0.000 000 001 n, nano
UNIT OF MEASUREMENTGRAVITATIONAL ATTRACTION
Exercise
Exercise
1. If a car is travelling at 88km/h, determine its speed in meter per seconds.
2. Evaluate each of the following and express with SI units having an appropriate prefixa. (50mN)(6GN)b. (400mm)(0.6 MN)²c. 45 MN³/900Gg
3. Two particles have a mass of 8kg and 12kg, respectively. If they are 800mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.
4. Use Newton’s law of universal gravitation to calculate the weight of a 70kg person standing on the surface of the earth. Then repeat the calculation by using W=mg and compare your two results. (Given R= 6371100, me= 5.976 X10^24)
4. Scalars & Vectors
2 kinds of quantities in mechanics- Scalars & Vectors
Scalar quantities are those with which only a magnitude is associated. I.e. time, volume, density, speed, energy, mass
Vector quantities possesses direction as well as magnitude, and must obey the parallelogram law of addition. I.e. displacement, velocity, acceleration, force, moment, and momentum
For ex: Speed is a scalar. It is the magnitude of velocity, which is a vector
4. Scalars & Vectors
Vectors representing physical quantities can be classified as;
Free vector is one whose action is not confined to a unique line in space. I.e. if a body moves without rotation, then the movement of any point in the body may be taken as vector. This vector describes equally well the direction and magnitude of the displacement of every point in the body. Thus, we may represent the displacement of such a body by a free vector
Sliding vector has a unique line of action is space but not a unique point of application. i.e. when an external force acts on a rigid body, the force can be applied at any point along its line of action without changing its effect on the body as a whole.
Fixed vector is one for which a unique point of application is specified. The action of a force on a deformable body must be specified by a fixed vector at the point of application of the force.
4. Scalars & Vectors
The length of the arrow represents the magnitude of the vector
The angle θ between the vector and a fixed axis defines the direction of its line of action.
The head of the arrow indicates the sense of direction of the vector
-V
V
θ
In scalar equation, the symbol will appear in lightface italic type, V.
Boldface type is used for vector quantities, V
When writing vector equation, always be certain to preserve the mathematical distinction between vectors and scalars. Use distinguishing mark for each vector quantity, I.e. V or V to take the place of boldface type in print.
4. Scalars & Vectors
4. Scalars & Vectors
Vector must obey the parallelogram law of addition.
V1 and V2 treated as free vectors, may be replaced by their equivalent vector V, which is diagonal of the parallelogram formed by V1 and V2 as its two sides.
The combination is called the vector sum V = V1 + V2 (vector) V ≠ V1 + V2 (scalar)
Parallelogram can then be reduced to a triangle, which represents the triangle rule
4. Scalars & Vectors
The two vectors V1 and V2 again treated as free vectors, may also be added head-to-tail by the triangle law to obtain the identical vector sum V
The difference V1 – V2 is easily obtained by adding –V2 to V1, where either the triangle or parallelogram
procedure may be used.V’ = V1 – V2, where the minus sign denotes vector
subtraction
4. Scalars & Vectors
Problems that involve the addition of two forces can be solved as follows:
Step 1. Parallelogram LawStep 2. TrigonometryCosine law:
C= √A² + B² - 2AB cos C
Sine Law: A/ sin a = B/sin b = c/sin c
4. Scalars & Vectors
When a force is resolved into two components along the x and y axes, the components are then called ‘Rectangular components’
Then,
Example 1
Figure below show two position vectors, the magnitudes of which are A=60m, and B=100m. A position vector is a vector drawn between two points in space. Determine the resultant R=A+B using the following methods
(1) analytically, using triangle law(2) graphically, using the triangle law
30 º 70 º
AB
Example 2
The vertical force P of magnitude 100kN is applied to the frame shown in figure below. Resolve P into components that are parallel to the members AB and AC of the truss
A
P
B
C
70º
35º
Example 3
4.2 Scalar notation
The rectangular components of force V shown in figure below are found using the parallelogram law, so that V= Vx + Vy
Because these components form a right triangle, their magnitudes can be determined from
Vx = Vcos θVy = V sin θ
4.3 Cartesian Vector Notation
It is also possible to represent the x and y components of a force in terms of Cartesian unit vectors i and j
Each of these unit vectors has a dimensionless magnitude of one, and so they can be used to designate the directions of the x and y axes
V = Vx i + Vy j
4.4 Coplanar Force Resultants
We can use either Cartesian Vector Notation or the Scalar notation to determine the resultant of several coplanar forces
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4.4 Coplanar Forces
Scalar Notation x and y axes are designated positive and
negative Components of forces expressed as
algebraic scalars
sin and cos FFFF
FFF
yx
yx
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4.4 Coplanar Forces
Cartesian Vector Notation Cartesian unit vectors i and j are used to
designate the x and y directions Unit vectors i and j have dimensionless
magnitude of unity ( = 1 ) Magnitude is always a positive
quantity, represented by scalars Fx and Fy
jFiFF yx
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4.4 Coplanar Forces
Coplanar Force ResultantsTo determine resultant of several coplanar
forces: Resolve force into x and y components Addition of the respective components
using scalar algebra Resultant force is found using the
parallelogram law Cartesian vector notation:
jFiFF
jFiFF
jFiFF
yx
yx
yx
333
222
111
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4.4 Coplanar Forces
Coplanar Force Resultants Vector resultant is therefore
If scalar notation are used
jFiF
FFFF
RyRx
R
321
yyyRy
xxxRx
FFFF
FFFF
321
321
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4.4 Coplanar Forces
Coplanar Force Resultants In all cases we have
Magnitude of FR can be found by Pythagorean Theorem
yRy
xRx
FF
FF
Rx
RyRyRxR F
FFFF 1-22 tan and
* Take note of sign conventions
Example 4
Using method (1)Scalar notation(2)Cartesian Vector Notation
Example 5
Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force RP=800 N
T= 600 N
α
A C D
B
60°
6m
3m
Using - Graphical Method, - Geometric method, - Algebraic method (xy coordinate)
Example 6
Using method (1)Scalar notation(2)Cartesian Vector Notation
Example 7
5. Cartesian Vectors
In particularly 3-D problems, it is convenient to express the rectangular components of V in terms of unit vectors i , j , k, which are vector in x-, y- and z-direction.
Cartesian Vector Separating magnitude and direction Simplify vector algebra
V = Vx + Vy + Vz
5. Cartesian Vectors
Magnitude of a Cartesian Vector
Direction of a Cartesian Vector Cos α = Vx/ V, Cos β = Vy/ V Cosγ = Vz/ V
An easy way of obtaining these direction cosines is to form a unit vector nv in the direction of V. nv will have a magnitude of one and be dimensionless provided V is divided by its magnitude.
5. Cartesian Vectors
A vector V may be expressed mathematically by multiplying its magnitude V by a vector n whose magnitude is one and whose direction coincides with that of V. The n is called a ‘unit vector’.
nv Represent the direction of cosines of V nv = Cos α i + Cos β j + Cosγ k Cos ²α + Cos ²β + Cos²γ = 1
n = V/ V
Example 8
Objective 2
To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction.
2 D 3 D
Example 9
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Solution I
Scalar Notation:
N
NNF
FF
N
NNF
FF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600
:
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Solution I
Resultant Force
From vector addition, direction angle θ is
N
NNFR629
8.5828.236 22
9.67
8.236
8.582tan 1
N
N
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Solution II
Cartesian Vector NotationF1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus, FR = F1 + F2
= (600cos30ºN - 400sin45ºN)i + (600sin30ºN + 400cos45ºN)j = {236.8i + 582.8j}NThe magnitude and direction of FR are determined in the same manner as before.
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2.5 Cartesian Vectors
Right-Handed Coordinate SystemA rectangular or Cartesian coordinate system is said to be right-handed provided: Thumb of right hand points in the direction
of the positive z axis z-axis for the 2D problem would be
perpendicular, directed out of the page.
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2.5 Cartesian Vectors
Rectangular Components of a Vector A vector A may have one, two or three
rectangular components along the x, y and z axes, depending on orientation
By two successive application of the parallelogram lawA = A’ + Az
A’ = Ax + Ay
Combing the equations, A can be expressed asA = Ax + Ay + Az
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2.5 Cartesian Vectors
Unit Vector Direction of A can be specified using a unit
vector Unit vector has a magnitude of 1 If A is a vector having a magnitude of A ≠ 0,
unit vector having the same direction as A is expressed by uA = A / A. So that
A = A uA
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2.5 Cartesian Vectors
Cartesian Vector Representations 3 components of A act in the positive i, j and
k directions
A = Axi + Ayj + AZk
*Note the magnitude and direction of each components are separated, easing vector algebraic operations.
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2.5 Cartesian Vectors
Magnitude of a Cartesian Vector From the colored triangle,
From the shaded triangle,
Combining the equations gives magnitude of A
222zyx AAAA ++=
22' yx AAA +=
22' zAAA +=
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2.5 Cartesian Vectors
Direction of a Cartesian Vector Orientation of A is defined as the coordinate
direction angles α, β and γ measured between the tail of A and the positive x, y and z axes
0° ≤ α, β and γ ≤ 180 ° The direction cosines of A is
A
Axcos
A
Aycos
A
Azcos
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2.5 Cartesian Vectors
Direction of a Cartesian Vector Angles α, β and γ can be determined by the
inverse cosinesGiven A = Axi + Ayj + AZk
then, uA = A /A = (Ax/A)i + (Ay/A)j + (AZ/A)k
where 222zyx AAAA
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2.5 Cartesian Vectors
Direction of a Cartesian Vector uA can also be expressed as
uA = cosαi + cosβj + cosγk
Since and uA = 1, we have
A as expressed in Cartesian vector form isA = AuA = Acosαi + Acosβj + Acosγk = Axi + Ayj + AZk
222zyx AAAA
1coscoscos 222
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2.6 Addition and Subtraction of Cartesian Vectors
Concurrent Force Systems Force resultant is the vector sum of all the
forces in the system
FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk
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Example 10
Express the force F as Cartesian vector.
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Solution
Since two angles are specified, the third angle is found by
Two possibilities exit, namely
1205.0cos 1
( ) o605.0cos 1 == -a
( ) ( )
oo
5.0707.05.01cos
145cos60coscos
1coscoscos
22
222
222
±=--=
=++
=++
a
a
gba
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Solution
By inspection, α = 60º since Fx is in the +x directionGiven F = 200N
F = Fcosαi + Fcosβj + Fcosγk = (200cos60ºN)i + (200cos60ºN)j
+ (200cos45ºN)k = {100.0i + 100.0j + 141.4k}N
Checking:
N
FFFF zyx
2004.1410.1000.100 222
222
Example 10
Note: ‘+ve’ or ‘-ve’ sign MUST take into account when
substituting into Cos α, cos β and cos γ
Example 11
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2.7 Position Vectors
x,y,z Coordinates Right-handed coordinate system Positive z axis points upwards, measuring the
height of an object or the altitude of a point Points are measured relative
to the origin, O.
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2.7 Position Vectors
Position Vector Position vector r is defined as a fixed vector
which locates a point in space relative to another point.
E.g. r = xi + yj + zk
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2.7 Position Vectors
Position Vector Vector addition gives rA + r = rB
Solving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)kor r = (xB – xA)i + (yB – yA)j + (zB –zA)k
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2.7 Position Vectors
Length and direction of cable AB can be found by measuring A and B using the x, y, z axes
Position vector r can be established Magnitude r represent the length of cable Angles, α, β and γ represent the direction
of the cable Unit vector, u = r/r
Example 12
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Solution
Position vectorr = [-2m – 1m]i + [2m – 0]j + [3m – (-
3m)]k = {-3i + 2j + 6k}m
Magnitude = length of the rubber band
Unit vector in the director of ru = r /r = -3/7i + 2/7j + 6/7k
mr 7623 222
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Solution
α = cos-1(-3/7) = 115°
β = cos-1(2/7) = 73.4°
γ = cos-1(6/7) = 31.0°
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2.8 Force Vector Directed along a Line
In 3D problems, direction of F is specified by 2 points, through which its line of action lies
F can be formulated as a Cartesian vector
F = F u = F (r/r)
Note that F has units of forces (N) unlike r, with units of length (m)
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2.8 Force Vector Directed along a Line
Force F acting along the chain can be presented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length of chain
Unit vector, u = r/r that defines the direction of both the chain and the force
We get F = Fu
Example 13
Example 14